Hydrostatic force
- Pressure at depth
- p = ρgh
- Force on a vertical wall
- F = ρg·hc·A
- Centre of pressure
- yp = yc + Ic/(ycA)
- Rectangle
- Ic = bH³/12
Hydraulic Engineering: cram pack
Formula sheets, the key method steps for every topic and 10 practice problems, with full solutions at the back.
1Formulas and methods
Continuity & Bernoulli
- Continuity
- A1v1 = A2v2
- Bernoulli
- p1 + ½ρv1² + ρgz1 = p2 + ½ρv2² + ρgz2
- Flow rate
- Q = A·v
- Circle area
- A = πd²/4
Pipe friction & head loss
- Reynolds number
- Re = V·D/ν = ρ·V·D/μ
- Flow regime
- laminar below Re ≈ 2000, turbulent above Re ≈ 4000
- Mean velocity
- V = Q/A, A = π·D²/4
- Darcy–Weisbach
- hf = f·(L/D)·V²/(2g)
- Minor losses
- hm = ΣK·V²/(2g)
- Typical K values
- entrance 0.5, exit 1.0, 90° bend ≈ 0.3
- Pressure drop
- Δp = ρ·g·hf
- Two open reservoirs
- Δz = (f·L/D + ΣK)·V²/(2g)
- Free outlet
- H = (1 + f·L/D + ΣK)·V²/(2g)
Pumps & pump power
- Hydraulic power
- Ph = ρ·g·Q·H = Q·Δp
- Shaft power
- Pshaft = Ph/ηpump
- Electrical power
- Pel = Pshaft/ηmotor
- Pump head
- H = Δz + Δp/(ρ·g) + hL
- System losses
- hL = (f·L/D + ΣK)·V²/(2g)
Open-channel flow (Manning)
- Manning equation
- V = (1/n)·R2/3·S1/2
- Flow rate
- Q = A·V
- Hydraulic radius
- R = A/P (P = wetted perimeter, free surface excluded)
- Rectangle
- A = b·y, P = b + 2y
- Trapezoid (side slope z:1)
- A = (b + z·y)·y, P = b + 2y·√(1 + z²)
- Circular pipe, full or half full
- R = D/4
- Required slope
- S = (n·Q/(A·R2/3))²
2Practice problems
1.Hydrostatic forceMedium
A vertical rectangular gate 1 m wide sits in the wall of a tank of water, with its top edge 2 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 2 mH = 1.5 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
2.Hydrostatic forceHard
A vertical rectangular gate 1 m wide sits in the wall of a tank of water, with its top edge 1 m below the surface. Find the resultant hydrostatic force F on the gate.
d = 1 mH = 3 mb = 1 mρ = 1000 kg/m³g = 9.81 m/s²
F = kN
3.Continuity & BernoulliMedium
Water flows through a horizontal pipe that narrows from 150 mm to 40 mm. In the wide section the pressure is 140 kPa and the velocity is 1 m/s. Ignoring losses, find the pressure p2 in the narrow section.
d1 = 150 mmd2 = 40 mmp1 = 140 kPav1 = 1 m/sρ = 1000 kg/m³
p2 = kPa
4.Continuity & BernoulliHard
Water flows up through a pipe that narrows from 150 mm to 75 mm while rising 2 m. At section 1 the pressure is 320 kPa and the velocity is 3 m/s. Ignoring losses, find the pressure p2 at section 2.
d1 = 150 mmd2 = 75 mmp1 = 320 kPav1 = 3 m/sΔz = 2 mρ = 1000 kg/m³
p2 = kPa
5.Pipe friction & head lossMedium
Water (ρ = 1000 kg/m³) flows at 1.2 m/s through a horizontal pipe, 800 m long, with an inside diameter of 75 mm. Use the Darcy friction factor f = 0.02 and g = 9.81 m/s². Find the pressure drop Δp over the pipe in kPa.
V = 1.2 m/sL = 800 mD = 75 mmf = 0.02ρ = 1000 kg/m³
Δp = kPa
6.Pipe friction & head lossHard
Two open reservoirs are joined by a pipe 100 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.02. The minor losses are a square-edged entrance (K = 0.5), 2 bends (K = 0.3 each), a valve (K = 2) and the exit into the lower reservoir (K = 1.0). Use g = 9.81 m/s². What difference in water level Δz between the reservoirs gives a flow rate of 40 L/s?
Q = 40 L/sL = 100 mD = 150 mmf = 0.02Kentrance = 0.5Kbend = 0.3 (×2)Kvalve = 2Kexit = 1.0
Δz = m
7.Pumps & pump powerMedium
A pump delivers 20 L/s of water (ρ = 1000 kg/m³) against a head of 66 m. The pump efficiency is 60% and the electric motor driving it has an efficiency of 88%. Use g = 9.81 m/s². Find the electrical power Pel the motor draws, in kW.
Q = 20 L/sH = 66 mηpump = 60%ηmotor = 88%g = 9.81 m/s²
Pel = kW
8.Pumps & pump powerHard
A pump moves 30 L/s of water (ρ = 1000 kg/m³) from an open lower reservoir to an upper tank whose water surface is 8 m higher. The pipe is 300 m long with an inside diameter of 150 mm and a Darcy friction factor f = 0.025; the minor-loss coefficients add up to ΣK = 6 (entrance and exit included). Use g = 9.81 m/s². The pump efficiency is 78%. Find the shaft power Pshaft in kW.
Q = 30 L/sΔz = 8 mL = 300 mD = 150 mmf = 0.025ΣK = 6η = 78%
Pshaft = kW
9.Open-channel flow (Manning)Medium
A trapezoidal finished concrete channel has a bottom width of 2 m and side slopes of 1.5:1 (horizontal:vertical). Water flows uniformly 1.7 m deep. Manning n = 0.012 and the bed slope is S = 0.001. Find the flow rate Q in m³/s.
b = 2 my = 1.7 mz = 1.5n = 0.012S = 0.001
Q = m³/s
10.Open-channel flow (Manning)Hard
A rectangular unfinished concrete channel 1 m wide must carry 1.4 m³/s in uniform flow at a depth of 0.7 m. Manning n = 0.014. Find the required bed slope S (in m/m).
Q = 1.4 m³/sb = 1 my = 0.7 mn = 0.014
S = m/m
3Solutions
1.Hydrostatic force
Answer: F = 40.47 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 2 + 1.5/2 = 2.75 m
A = H·b = 1.5·1 = 1.5 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.75·1.5 = 40.47 kN
Answer
F = 40.47 kN
2.Hydrostatic force
Answer: F = 73.58 kN
Hydrostatic force on a plane surface: F = ρ·g·yc·A. It acts at the centre of pressure, ycp = yc + Ic/(ycA), which always lies below the centroid.
Centroid depth and area
yc = d + H/2 = 1 + 3/2 = 2.5 m
A = H·b = 3·1 = 3 m²
Resultant force
F = ρ·g·yc·A = 1000·9.81·2.5·3 = 73.58 kN
Answer
F = 73.58 kN
3.Continuity & Bernoulli
Answer: p2 = 41.62 kPa
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Continuity and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 1·(150/40)² = 14.06 m/s
½ρ(v1² − v2²) = ½·1000·(1² − 14.06²)/1000 = −98.38 kPa
p2 = p1 + ½ρ(v1² − v2²) = 140 + (−98.38) = 41.62 kPa
Answer
p2 = 41.62 kPa
4.Continuity & Bernoulli
Answer: p2 = 232.9 kPa
Continuity and Bernoulli: for an incompressible fluid the flow rate is the same at every section, Q = A₁v₁ = A₂v₂. Along a streamline without losses, p + ½ρv² + ρgz is constant, so where the fluid is faster or higher its pressure is lower.
Continuity and Bernoulli
Work in Pa, then divide by 1000 to get kPa.
v2 = v1·(d1/d2)² = 3·(150/75)² = 12 m/s
½ρ(v1² − v2²) = ½·1000·(3² − 12²)/1000 = −67.5 kPa
ρgΔz = 1000·9.81·2/1000 = 19.62 kPa
p2 = p1 + ½ρ(v1² − v2²) − ρgΔz = 320 + (−67.5) − 19.62 = 232.9 kPa
Answer
p2 = 232.9 kPa
5.Pipe friction & head loss
Answer: Δp = 153.6 kPa
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Head loss
hf = f·(L/D)·V²/(2g) = 0.02·(800/0.075)·1.2²/(2·9.81) = 15.66 m
Pressure drop
The pipe is horizontal and the velocity does not change, so the energy equation reduces to Δp = ρ·g·hf.
Δp = ρ·g·hf = 1000·9.81·15.66 = 153600 Pa = 153.6 kPa
Answer
Δp = 153.6 kPa
6.Pipe friction & head loss
Answer: Δz = 4.553 m
Pipe flow: the Reynolds number Re = V·D/ν = ρ·V·D/μ tells laminar flow (Re below about 2000) from turbulent flow (above about 4000). Friction along a pipe costs the Darcy–Weisbach head loss hf = f·(L/D)·V²/(2g). Fittings, valves, the entrance and the exit add minor losses hm = ΣK·V²/(2g). Between two open reservoir surfaces the energy equation gives Δz = hf + hm, with the exit loss K = 1 included.
Velocity
V = Q/A = 0.04/0.01767 = 2.264 m/s
V²/(2g) = 0.2611 m
Loss coefficients
f·L/D = 0.02·100/0.15 = 13.33
ΣK = 0.5 + 2·0.3 + 2 + 1.0 = 4.1
Total head loss
hf = 13.33·0.2611 = 3.482 m
hm = 4.1·0.2611 = 1.071 m
Δz = hf + hm = 4.553 m
Answer
Δz = 4.553 m
7.Pumps & pump power
Answer: Pel = 24.53 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Hydraulic power
Ph = ρ·g·Q·H = 1000·9.81·0.02·66 = 12.95 kW
Shaft power
Pshaft = 12.95/0.6 = 21.58 kW
Electrical power
Pel = 21.58/0.88 = 24.53 kW
Answer
Pel = 24.53 kW
8.Pumps & pump power
Answer: Pshaft = 6.122 kW
Pump power: the power given to the water is the hydraulic power Ph = ρ·g·Q·H = Q·Δp. The shaft must deliver more because the pump loses some: Pshaft = Ph/ηpump, and the motor draws Pel = Pshaft/ηmotor. The pump head H follows from the energy equation: static lift + pressure head + all friction and minor losses.
Velocity
V = Q/A = 0.03/0.01767 = 1.698 m/s
V²/(2g) = 0.1469 m
Head losses
hL = (f·L/D + ΣK)·V²/(2g) = (50 + 6)·0.1469 = 8.226 m
Pump head
H = Δz + hL = 8 + 8.226 = 16.23 m
Shaft power
Pshaft = ρ·g·Q·H/η = 1000·9.81·0.03·16.23/0.78 = 6.122 kW
Answer
Pshaft = 6.122 kW
9.Open-channel flow (Manning)
Answer: Q = 19.72 m³/s
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
Geometry
A = (b + z·y)·y = (2 + 1.5·1.7)·1.7 = 7.735 m²
P = b + 2y·√(1 + z²) = 2 + 2·1.7·√(1 + 1.5²) = 8.129 m
R = A/P = 0.9515 m
Velocity
V = (1/0.012)·0.95152/3·0.0011/2 = 2.549 m/s
Flow rate
Q = A·V = 7.735·2.549 = 19.72 m³/s
Answer
Q = 19.72 m³/s
10.Open-channel flow (Manning)
Answer: S = 0.004053 m/m
Manning’s equation (SI): for uniform flow in an open channel V = (1/n)·R2/3·S1/2 and Q = A·V, with n the roughness coefficient, S the bed slope (m/m) and R = A/P the hydraulic radius: flow area divided by wetted perimeter. Only the boundary in contact with the water counts in P, not the free surface. Rectangle: A = b·y, P = b + 2y. Trapezoid with side slopes z:1 (horizontal:vertical): A = (b + z·y)·y, P = b + 2y·√(1 + z²). A circular pipe flowing full or half full has R = D/4.
Geometry
A = b·y = 1·0.7 = 0.7 m²
P = b + 2y = 1 + 2·0.7 = 2.4 m
R = A/P = 0.2917 m
Solve Manning for S
S1/2 = n·Q/(A·R2/3) = 0.014·1.4/(0.7·0.4398) = 0.06366
S = 0.06366² = 0.004053
Answer
S = 0.004053
That is a fall of 4.053 m per kilometre.
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